What percentage of job opening are published?

a. 10% - 15%

b. 15% - 20%

30% - 35%

35% - 40%

Please select the best answer from the choices provided

Ο

Α

Answer:

6.67% probability that the result is a multiple of 5 and a multiple of 2

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

Desired outcomes:

Multiples of 2 AND 5

Between 1 and 15, the multiples of 2 are: 2,4,6,8,10,12,14

Between 1 and 15, the multiples of 5 are: 5,10,15

So only 10 is a multiply of both 2 and 5, so only one desired outcome, which means that [tex]D = 1[/tex]

Total outcomes:

Any number between 1 and 15, there are 15, so [tex]T = 15[/tex]

Probability:

[tex]p = \frac{D}{T} = \frac{1}{15} = 0.0667[/tex]

6.67% probability that the result is a multiple of 5 and a multiple of 2

Answer:

Step-by-step explanation:

3/5 or 9/15

Answer:

The value of the sample mean is 78.

Step-by-step explanation:

We are given that a sample of n = 4 scores is obtained from a population with a mean of 70 and a standard deviation of 8.

Also, the sample mean corresponds to a z score of 2.00.

Let [tex]\bar X[/tex] = sample mean

The z-score probability distribution for a sample mean is given by;

              Z = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]  ~ N(0,1)

where, [tex]\mu[/tex] = population mean = 70

            [tex]\sigma[/tex] = standard deviation = 8

            n = sample size = 4

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, we are given that the sample mean corresponds to a z score of 2.00 for which we have to find the value of sample mean;

So, z-score formula is given by ;

                  z-score = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] = [tex]\frac{\bar X-70}{\frac{8}{\sqrt{4} } }[/tex]

                      2.00 =  [tex]\frac{\bar X-70}{\frac{8}{\sqrt{4} } }[/tex]

                      2.00 =  [tex]\frac{\bar X-70}{4 } }[/tex]

                     [tex]\bar X = 70+(2 \times 4)[/tex]

                     [tex]\bar X[/tex] = 70 + 8 = 78

Therefore, the value of the sample mean is 78.

The value of the sample mean is 78.

To find the value of the sample mean, we can use the formula:

sample mean = population mean + (z score * (standard deviation / square root of sample size))

In this case, the population mean is 70, the z score is 2.00, and the standard deviation is 8. Since the sample size is 4, we calculate the square root of 4, which is 2. Plugging these values into the formula gives us:

sample mean = 70 + (2.00 * (8 / 2)) = 70 + (2.00 * 4) = 70 + 8 = 78...

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Answer:

After 15 years, the area will be of 775.3 km²

Step-by-step explanation:

The equation for the area of the forest after t years has the following format.

[tex]A(t) = A(0)(1-r)^{t}[/tex]

In which A(0) is the initial area and r is the yearly decrease rate.

A certain forest covers an area of 2600 km2.

This means that [tex]A(0) = 2600[/tex]

Suppose that each year this area decreases by 7.75%.

This means that [tex]r = 0.0775[/tex]

So

[tex]A(t) = 2600(1-0.0775)^{t}[/tex]

[tex]A(t) = 2600(0.9225)^{t}[/tex]

What will the area be like after 15 years?

This is [tex]A(15)[/tex]

[tex]A(t) = 2600(0.9225)^{t}[/tex]

[tex]A(15) = 2600(0.9225)^{15} = 775.3[/tex]

After 15 years, the area will be of 775.3 km²

Answer:

[tex] A(t) = 2600 (1-0.0775)^t = 2600 (0.9225)^t [/tex]

And since the question wants the value for the area at t = 15 years from know we just need to replace t=15 in oir model and we got:

[tex] A(15) = 2600 (0.9225)^{15} = 775.299[/tex]

So then we expect about 775.299 km2 remaining for the area of forests.

Step-by-step explanation:

For this case we can use the following model to describe the situation:

[tex] A = A_o (1 \pm r)^{t}[/tex]

Where [tex]A_o = 2600 km^2[/tex] represent the initial area

[tex] r =-0.0775[/tex] represent the decreasing rate on fraction

A represent the amount of area remaining and t the number of years

So then our model would be:

[tex] A(t) = 2600 (1-0.0775)^t = 2600 (0.9225)^t [/tex]

And since the question wants the value for the area at t = 15 years from know we just need to replace t=15 in oir model and we got:

[tex] A(15) = 2600 (0.9225)^{15} = 775.299[/tex]

So then we expect about 775.299 km2 remaining for the area of forests.

Answer:

The salmon size is less than equal to 10.28 inches represent bottom 25% of all salmon.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 12.5 inches

Standard Deviation, σ = 3.3

We are given that the distribution of salmon size is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

We have to find the value of x such that the probability is 0.25

[tex]P( X < x) = P( z < \displaystyle\frac{x - 12.5}{3.3})=0.25[/tex]  

Calculation the value from standard normal z table, we have,  

[tex]\displaystyle\frac{x - 12.5}{3.3} = -0.674\\\\x = 10.2758\approx 10.28[/tex]  

Thus, the salmon size is equal to or less than 10.28 inches, they are considered small and young and represent bottom 25% of all salmon.

Answer:

$371.39

Step-by-step explanation:

150 * .12+150 = 168

168 * .12+168 = 188.16

188.16 * .12+188.16 = 210.7392

210.7392 * .12+210.7392 = 236.027904

264.3512525 * .12+264.3512525 = 296.0734028

296.0734028 * .12+296.0734028 = 331.6022111

331.6022111 * .12+331.6022111 = 371.3944764

you multiply your current number by 12% and add that to the number, the last number i rounded for the answer as you can see

Answer:

  yes

Step-by-step explanation:

The 90-th percentile of a normal distribution corresponds to a z-score of 1.282. Bob's z-score is above that, so he is definitely in the top 10%.

_____

If you haven't memorized the percentiles associated with the normal distribution, it is convenient to use a calculator or table of values.

In this case, ob is correct in saying that his score is in the top 10 percent.

To determine if Bob's z-score corresponds to a score in the top 10 percent of the distribution, we need to find the percentile associated with his z-score.

We can then compare this percentile to 90%, as the top 10% corresponds to the highest scores.

Using a standard normal distribution table or a calculator, we find that a z-score of 1.52 corresponds approximately to the 93rd percentile.

This means that 93% of the scores are below Bob's score, indicating that he is indeed in the top 10% of the class.

So, Bob is correct in saying that his score is in the top 10 percent.

Answer:

5

Step-by-step explanation:

It be 5 because its in the middle

3,4,7,8

Step-by-step explanation:

1 2 3 4 5 6 7 8 9

3 4 7 8

Answer: £569.40p

Step-by-step explanation:

410÷£0.72=£569.40p

Answer:

£295.20

Step-by-step explanation:

€410 x 0.72 = £295.20

Answer:

[tex]15.55-2.997\frac{0.278}{\sqrt{8}}=15.26[/tex]    

[tex]15.55+2.997\frac{0.278}{\sqrt{8}}=15.84[/tex]    

The 98% confidence interval would be given by (15.26;15.84)    

Step-by-step explanation:

Notation

[tex]\bar X[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

We can calculate the mean and the sample deviation we can use the following formulas:  

[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)  

[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)  

The mean calculated for this case is [tex]\bar X=15.55[/tex]

The sample deviation calculated [tex]s=0.278[/tex]

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=8-1=7[/tex]

Since the Confidence is 0.98 or 98%, the value of [tex]\alpha=0.02[/tex] and [tex]\alpha/2 =0.01[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.01,7)".And we see that [tex]t_{\alpha/2}=2.997[/tex]

And the confidence interval is given by:

[tex]15.55-2.997\frac{0.278}{\sqrt{8}}=15.26[/tex]    

[tex]15.55+2.997\frac{0.278}{\sqrt{8}}=15.84[/tex]    

The 98% confidence interval would be given by (15.26;15.84)    

Answer:

The base of the triangle = 9inches

Step-by-step explanation:

Explanation:-

The area of the triangle = [tex]\frac{1}{2} bh[/tex] square units

Given area of the triangle(A) = 54 square inches

Given the height of the triangle (h) = 12 inches

now  equating     [tex]\frac{1}{2} b h = 54[/tex]

                          [tex]\frac{1}{2} b (12) = 54[/tex]

now simplification, we get

                           6 b = 54

Dividing '6' on both sides, we get

                             b = 9

The base of the triangle = 9 inches

Conclusion:-

The base of the triangle = 9 inches

Answer:

C) 192.11

Step-by-step explanation:

To find the mean of any group of numbers, you simply add up all the numbers and then divide by how many numbers there are.

Using a calculator or a pencil, add the following:

180 + 220 + 240 + 196 + 175 + 183 + 195 + 140 + 200

= 1,729

The last step is to divide 1,729 by 9:

1,729 ÷ 9 = 192.11

Answer:

Cube surface area = 6 * side^2

150 square inches = 6 * side^2

sq root (side) = 25

side = 5 inches

Step-by-step explanation:

Answer:

555

Step-by-step explanation:

Answer:

the edges only would be 16 pieces.

Step-by-step explanation:

Think of it as a square. The top row is 5 pieces, bottom row is 5 pieces.

The left vertical side is 3 and the right vertical side is 3. Because the corner pieces are already there, you don't count them twice.

Answer:

The weight of an adult male who is 158 cm tall is 59.32 kg.

Step-by-step explanation:

The regression equation representing the relationship between height and weight of a person is:

[tex]\hat y=-105+1.04 x[/tex]

Here,

y = weight of a person (in kg)

x = height of a person (in cm)

The information provided is:

[tex]\bar x=167.77\ \text{cm}\\\bar y=81.54\ \text{cm}\\r(X, Y)=0.201\\p-value=0.045[/tex]

The significance level of the test is, α = 0.01.

The hypothesis to test the significance of the correlation between height and weight is:

H₀: There is no relationship between the height and weight, i.e. ρ = 0.

Hₐ: There is a relationship between the height and weight, i.e. ρ ≠ 0.

Decision rule:

If the p-value of the test is less than the significance level, then the null hypothesis will be rejected and vice-versa.

According to information provided:

p-value = 0.045 > 0.01

The null hypothesis will not be rejected at 1% level of significance.

Thus, concluding that there is no relationship between the height and weight.

Compute the weight of an adult male with height, x = 158 cm as follows:

[tex]\hat y=-105+1.04 x[/tex]

  [tex]=-105+(1.04\times 158)\\=-105+164.32\\=59.32[/tex]

Thus, the weight of an adult male who is 158 cm tall is 59.32 kg.

Answer:

P(A/D) = 0.5714

Step-by-step explanation:

Let's call A the event that a chip is produced by Plant A, B the event that a chip is produced by Plant B and D the event that the chip is defective

So, the likelihood or probability P(A/D) that a chip came from plant A given that the chip is defective is calculated as:

P(A/D) = P(A∩D)/P(D)

Where P(D) = P(A∩D) + P(B∩D)

Then, the probability P(A∩D) that a chip is produced by plant A and it is defective is calculated as:

P(A∩D) = 0.4*0.02 = 0.008

Because, Plant A produces 40% of the chips and 2% of the chips produced by plant A are defective.

At the same way, the probability P(B∩D) that a chip is produced by plant B and it is defective is calculated as:

P(B∩D) = 0.6*0.01 = 0.006

So, P(D) and P(A/D) are equal to:

P(D) = 0.008 + 0.006 = 0.014

P(A/D) = 0.008/0.014 = 0.5714

it means that if a randomly chosen chip produced by the company is defective, the likelihood that the chip came from plant A is 0.5714

Answer: Top Right, A rectangle has all the properties of a square.

Step-by-step explanation: Nobody how hard you try to make a rectangle a square, it won't work. A rectangle cannot have 4 equal slides.

Answer:

120 cm

Step-by-step explanation:

10 plus 6 equals 16.

16 times 15 equals 240

240 divided by 2 equals 120

Answer:

120cm

Step-by-step explanation:

got it right on edge

Answer:

z = 1.960

Step-by-step explanation:

The sample proportion is:

p = 715 / 2684 = 0.2664

The standard error is:

σ = √(pq/n)

σ = √(0.266 × 0.734 / 2684)

σ = 0.0085

For α = 0.05, the confidence level is 95%.  The z-statistic at 95% confidence is 1.960.

The margin of error is 1.960 × 0.0085 = 0.0167.

The confidence interval is 0.2664 ± 0.0167 = (0.2497, 0.2831).

The upper limit is 28.3%, so the journal can conclude with 95% confidence that the true percentage is less than 29%.

Yes, the considered sample provides sufficient evidence to indicate that the true percentage of all firms that announced one or more acquisitions during the year 2000 is less than 29​%.

Suppose that we have:

Then, the z test statistic for one sample proportion is:

[tex]Z = \dfrac{\hat{p} - p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]

For this case, we're provided that:

We want to determine if true percentage of all firms that announced one or more acquisitions during the year 2000 is less than 29​% = 0.29 (converted percent to decimal).

Hypotheses:

Thus, the test is left tailed test.

where [tex]p_0[/tex] = 29% = 0.29 is the hypothesized mean value of population proportion.

The test statistic is:

[tex]Z = \dfrac{\hat{p} - p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\\\ Z = \dfrac{715/2684 - 0.29}{\sqrt{\dfrac{0.29(1-0.29)}{2684}}} \approx -2.69[/tex]

The critical value of Z at level of significance 0.05 is -1.6449

Since the test statistic = -2.69 < critical value = -1.6449, so the test statistic lies in the rejection region (the rejection region for the left tailed test is all the values below critical value).

Thus, we reject the null hypothesis and accept the alternative hypothesis that the true population mean is less than 0.29.

Thus, the considered sample provides sufficient evidence to indicate that the true percentage of all firms that announced one or more acquisitions during the year 2000 is less than 29​%.

The z-test statistic came out to be -2.69

Learn more about one-sample z-test for population proportion mean here:

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Answer:

1. y = f(8)

2. So for t which f'(t) > 0, the drug level in the bloodstream is increasing. And for t which f'(t) < 0, it is decreasing.

Step-by-step explanation:

The concentration of the drug in the bloodstream at t hours is:

y = f(t)

1. What is the concentration of the drug in the bloodstream at t= 8 hours?

At t hours, y = f(t)

So at 8 hours, y = f(8)

2. During what time interval is the drug level in the bloodstream increasing? Decreasing?

A function f(t) is increasing when

f'(t) > 0

And is decreasing when

f'(t) < 0

So for t which f'(t) > 0, the drug level in the bloodstream is increasing. And for t which f'(t) < 0, it is decreasing.

(1) The concentration  of drug in the bloodstream in 8 hours is given by

[tex]\rm \bold{y = f (8)}[/tex]

(2) The time interval for which [tex]\rm y'=f'(t)>0[/tex] the drug level of the bloodstream is increasing.

The time interval for which [tex]\rm y' = f'(t) <0[/tex] the drug level of the bloodstream is decreasing.

When a certain prescription drug is taken orally by an adult.

the amount of the drug (in mg/L) in the bloodstream at t hours is given by the function y=f(t)

To be determined

(1) The concentration of the drug in the bloodstream at t= 8 hours

(2) During what time interval is the drug level in the bloodstream increasing or deceasing

The amount of the drug (in mg/L) in the bloodstream at t hours is given by the function

y=f(t).......(1)

(1)  The concentration  of drug in the bloodstream in 8 hours is given by putting t= 8 in the equation (1) which can be formulated as below

[tex]\rm y = f (8)[/tex]

(2) From the definition of increasing and decreasing function we can write that

[tex]\rm y = f(x) \; is \; increasing \; when \; f' (x)>0 \\and\; y = f(x) ; is \; decreasing \; when \; f' (x)<0 \\\\\\\\[/tex]

By the definition of increasing and decreasing function we can say that

The time interval for which [tex]\rm y'=f'(t)>0[/tex] the drug level of the bloodstream is increasing.

The time interval for which [tex]\rm y'=f'(t) <0[/tex] the drug level of the blood stream is decreasing.

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Answer:

2) We are 95% confident that the difference between the average sales after the ad campaign and the average sales before the ad campaign is between-3.83 and 1.83

Step-by-step explanation:

The confidence interval is a estimation, from the information that the sample gives, about a parameter of the population. In this case, the difference of means.

The 95% is a measure of the confidence about the estimation about the difference of the means. There is a 95% probability that the difference of means (sales after and sales before the ad) is within the confidence interval.

Answer:

(a) P (Z < 2.36) = 0.9909                    (b) P (Z > 2.36) = 0.0091

(c) P (Z < -1.22) = 0.1112                      (d) P (1.13 < Z > 3.35)  = 0.1288

(e) P (-0.77< Z > -0.55)  = 0.0705       (f) P (Z > 3) = 0.0014

(g) P (Z > -3.28) = 0.9995                   (h) P (Z < 4.98) = 0.9999.

Step-by-step explanation:

Let us consider a random variable, [tex]X \sim N (\mu, \sigma^{2})[/tex], then [tex]Z=\frac{X-\mu}{\sigma}[/tex], is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, [tex]Z \sim N (0, 1)[/tex].

In statistics, a standardized score is the number of standard deviations an observation or data point is above the mean.  The z-scores are standardized scores.

The distribution of these z-scores is known as the standard normal distribution.

(a)

Compute the value of P (Z < 2.36) as follows:

P (Z < 2.36) = 0.99086

                   ≈ 0.9909

Thus, the value of P (Z < 2.36) is 0.9909.

(b)

Compute the value of P (Z > 2.36) as follows:

P (Z > 2.36) = 1 - P (Z < 2.36)

                   = 1 - 0.99086

                   = 0.00914

                   ≈ 0.0091

Thus, the value of P (Z > 2.36) is 0.0091.

(c)

Compute the value of P (Z < -1.22) as follows:

P (Z < -1.22) = 0.11123

                   ≈ 0.1112

Thus, the value of P (Z < -1.22) is 0.1112.

(d)

Compute the value of P (1.13 < Z > 3.35) as follows:

P (1.13 < Z > 3.35) = P (Z < 3.35) - P (Z < 1.13)

                            = 0.99960 - 0.87076

                            = 0.12884

                            ≈ 0.1288

Thus, the value of P (1.13 < Z > 3.35)  is 0.1288.

(e)

Compute the value of P (-0.77< Z > -0.55) as follows:

P (-0.77< Z > -0.55) = P (Z < -0.55) - P (Z < -0.77)

                                = 0.29116 - 0.22065

                                = 0.07051

                                ≈ 0.0705

Thus, the value of P (-0.77< Z > -0.55)  is 0.0705.

(f)

Compute the value of P (Z > 3) as follows:

P (Z > 3) = 1 - P (Z < 3)

             = 1 - 0.99865

             = 0.00135

             ≈ 0.0014

Thus, the value of P (Z > 3) is 0.0014.

(g)

Compute the value of P (Z > -3.28) as follows:

P (Z > -3.28) = P (Z < 3.28)

                    = 0.99948

                    ≈ 0.9995

Thus, the value of P (Z > -3.28) is 0.9995.

(h)

Compute the value of P (Z < 4.98) as follows:

P (Z < 4.98) = 0.99999

                   ≈ 0.9999

Thus, the value of P (Z < 4.98) is 0.9999.

**Use the z-table for the probabilities.

To find the probabilities, we use the standard normal distribution table or a calculator to calculate probabilities for a standard normal distribution. We calculate each probability step by step and round the answers to four decimal places.

To determine the probabilities, we will use the standard normal distribution table or a calculator that can calculate probabilities for a standard normal distribution.

(a) P(z < 2.36) = 0.9900

(b) P(z > 2.36) = 1 - P(z < 2.36) = 1 - 0.9900 = 0.0100

(c) P(z < -1.22) = 0.1103

(d) P(1.13 < z < 3.35) = P(z < 3.35) - P(z < 1.13) = 0.9993 - 0.8708 = 0.1285

(e) P(-0.77 < z < -0.55) = P(z < -0.55) - P(z < -0.77) = 0.2896 - 0.2823 = 0.0073

(f) P(z > 3) = 1 - P(z < 3) = 1 - 0.9987 = 0.0013

(g) P(z > -3.28) = 1 - P(z < -3.28) = 1 - 0.0005 = 0.9995

(h) P(z < 4.98) = 1 - P(z > 4.98) = 1 - 0.0000 = 1.0000

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Given:

Given that Brinson's family is going camping. Their tent is shaped like a rectangular pyramid.

The volume of the tent is 6000 cubic inches.

The area of the base of the tent is 1200 square inches.

We need to determine the height of the tent.

Height of the tent:

The height of the tent can be determined using the formula,

[tex]V=\frac{1}{3}Bh[/tex]

where B is the area of the base and h is the height of the pyramid.

Substituting V = 6000 and B = 1200, we get;

[tex]6000=\frac{1}{3}(1200)h[/tex]

[tex]6000=400h[/tex]

   [tex]15=h[/tex]

Thus, the height of the tent is 15 inches.

Answer:

Infinite pairs of numbers

1 and -1

8 and -8

Step-by-step explanation:

Let x³ and y³ be any two real numbers. If the sum of their cube roots is zero, then the following must be true:

[tex]\sqrt[3]{x^3}+ \sqrt[3]{y^3}=0\\ \sqrt[3]{x^3}=- \sqrt[3]{y^3}\\x=-y[/tex]

Therefore, any pair of numbers with same absolute value but different signs fit the description, which means that there are infinite pairs of possible numbers.

Examples: 1 and -1; 8 and -8; 27 and -27.

Answer:

This site is kinda sucky

Step-by-step explanation:

go to a different one

Answer:

1.6

Step-by-step explanation:

if you divide 16.8 by 4.5 you get 3.7 now 3.7 divided 2.3 = 1.6

c^2 = a^2 + b^2 - 2(ab)(cos C)

c^2 + 2(ab)(cos C) = a^2 + b^2

2(ab)(cos C) = a^2 + b^2 - c^2

cos C = (a^2 + b^2 - c^2) / 2ab - Answer choice E

Hope this helps! :)

To make COS the subject of a formula, the equation is rearranged such that COS is isolated. An example can be in the equation a = b cos(x) which can be rearranged as cos(x) = a/b. A complete formula is necessary for an accurate step-by-step guide.

To make COS the subject of a formula, it typically involves other known quantities represented by variables and constants. For example, in the equation a = b cos(x), we can make cos(x) the subject of the equation by rearranging it to: cos(x) = a/b. However, to provide a more accurate step-by-step guide, the complete formula is necessary. This principle can be applied to various trigonometric formulas so that COS becomes the main focus.

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Answer:

95% confidence interval for the actual proportion of Meridian Township residents who are in favor of the tax increase is [0.5228 , 0.6972].

Step-by-step explanation:

We are given that the Township Board of Meridian Township randomly surveyed a group of residents and found that 61% are in favor of the tax increase.

The estimated standard error of sample proportion is 0.0445.

Firstly, the pivotal quantity for 95% confidence interval for the population proportion is given by;

                          P.Q. = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex]  ~ N(0,1)

where, [tex]\hat p[/tex] = sample proportion of residents in favor of the tax increase = 61%

           n = sample of criminals

           p = population proportion

Here for constructing 99% confidence interval we have used One-sample z proportion test statistics.

The 95% confidence interval for the actual proportion of Meridian Township residents who are in favor of the tax increase is given by;

95% Confidence interval for p =  [tex]\hat p \pm Z_\frac{\alpha}{2} \times \sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex]

Here, Standard error =  [tex]\sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex] = 0.0445

And [tex]\alpha[/tex] = significance level

So,  [tex]Z_\frac{\alpha}{2} =Z_\frac{0.05}{2}[/tex] = 1.96

Hence, 95% Confidence interval for p =  [tex]0.61 \pm 1.96 \times 0.0445[/tex]

                                                           =  [[tex]0.61 -0.08722[/tex] , [tex]0.61 +0.08722[/tex]]

                                                           =  [0.5228 , 0.6972]

Therefore, 95% confidence interval for the actual proportion of Meridian Township residents who are in favor of the tax increase is [0.5228 , 0.6972].

Also, the upper bound for the 95% confidence interval is 0.6971.

Answer:

Step-by-step explanation:

(a) A parameter of a population measures the characteristics of the population, In the question the proportion of all the persons who have health insurance and the mean of the entire dollar amount that Americans spent on health care in the past year measure the population.

Invariably, the proportion of persons having health insurance, and the mean dollars spent on health care for all Americans are the population parameter

(b)  A statistics  measure the characteristics of the sample.

In the question, the sample of 1500 Americans are considered to estimate the proportion of all Americans, proportion of all the persons who have health insurance among 1500 and the sample mean of all the dollar amounts that the selected Americans spent on healthy care in the past year describe  the sample.

Invariably, the sample proportion of persons having health insurance, and the mean dollars spent on health care for 1500 selected Americans are sample statistics

I suppose you're supposed to prove that set intersection is distributive across a union,

[tex]A\cap(B\cup C)=(A\cap B)\cup(A\cap C)[/tex]

Two sets are equal if they are subsets of one another. To prove a set [tex]X[/tex] is a subset of another set [tex]Y[/tex], you have to show that any element [tex]x\in X[/tex] also belongs to [tex]Y[/tex].

Let [tex]x\in A\cap(B\cup C)[/tex]. By definition of intersection, both [tex]x\in A[/tex] and [tex]x\in B\cup C[/tex]. By definition of union, either [tex]x\in B[/tex] or [tex]x\in C[/tex]. If [tex]x\in B[/tex], then clearly [tex]x\in A\cap B[/tex]; if [tex]x\in C[/tex], then [tex]x\in A\cap C[/tex]. Either way, [tex]x\in(A\cap B)\cup(A\cap C)[/tex]. Hence [tex]A\cap(B\cup C)\subseteq(A\cap B)\cup(B\cap C)[/tex].

The proof in the other direction uses the same sort of reasoning. Let [tex]x\in(A\cap B)\cup(A\cap C)[/tex]. Then either [tex]x\in A\cap B[/tex] or [tex]x\in A\cap C[/tex]. If [tex]x\in A\cap B[/tex], then both [tex]x\in A[/tex] and [tex]x\in B[/tex]; if [tex]x\in A\cap C[/tex], then both [tex]x\in A[/tex] and [tex]x\in C[/tex]. So certainly [tex]x\in A[/tex], and either [tex]x\in B[/tex] or [tex]x\in C[/tex] so that [tex]x\in B\cup C[/tex]. Hence [tex](A\cap B)\cup(A\cap C)\subseteq A\cap(B\cup C)[/tex].

Both sets are subsets of one another, so they are equal.

Answer:

b. We can be 95% confident that the proportion of all nicotine patch users who would report no smoking incidents in the following year is between 17.4% and 24.8%.

Step-by-step explanation:

The confidence interval is an estimation for the true population parameter, calculated from the information of a sample of this population.

The parameter of the population will be within this interval with a certain degree of confidence.

a. There is a 95% probability that the proportion of all nicotine patch users who would report no smoking incidents in the following year is between 17.4% and 24.8%.

Incorrect. The confidence interval gives only the probability that the true proportion (or population proportion) is within 17.4% and 24.8%, not the proportion of individual samples.

b. We can be 95% confident that the proportion of all nicotine patch users who would report no smoking incidents in the following year is between 17.4% and 24.8%.

Correct.

c. We can be 95% confident that the proportion of the sample who would report no smoking incidents in the following year is between 17.4% and 24.8%.

Incorrect. The confidence interval does not give information about another samples.

d. 95% of samples will have between 17.4% and 24.8% who would report no smoking incidents in the following year.

Incorrect. The confidence interval does not give information about another samples or sampling distributions.